Question
Discuss the continuity and differentiability of,$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
✓
Answer
$\text{f(x)}=\begin{cases}(\text{x}-\text{c})\cos\Big(\frac{1}{\text{x}-\text{c}}\Big), & \text{x}\neq 0\\0, & \text{x}= 0\end{cases}$
(LHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
(RHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{c}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
f(c) = 0
Since, LHL = f(x) = RHL at x = c
⇒ f(x) is continuous at x = c
(LHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}-\text{h})-\text{f(c)}}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(RHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}+\text{h})-\text{f(c)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\cos\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(LHL at x = c) = (RHL at x = c)
So,
f(x) is differentiable and continuous at x = c.
(LHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{-}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
(RHL at x = c) $=\lim_\limits{\text{x}\rightarrow\text{c}^{+}}\text{f(x)}$
$=\lim_\limits{\text{h}\rightarrow0}\text{f}(\text{c}+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}-\text{h}\cos\Big(\frac{1}{\text{h}}\Big)$
$=0$
f(c) = 0
Since, LHL = f(x) = RHL at x = c
⇒ f(x) is continuous at x = c
(LHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}-\text{h})-\text{f(c)}}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}-\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}-\text{h}-\text{c}}\Big)-0}{-\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(-\frac{1}{\text{h}}\Big)$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(RHL at x = c) $=\lim_\limits{\text{h}\rightarrow0}\frac{\text{f}(\text{c}+\text{h})-\text{f(c)}}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{(\text{c}+\text{h}-\text{c})\cos\Big(\frac{1}{\text{c}+\text{h}-\text{c}}\Big)-0}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\text{h}\cos\Big(\frac{1}{\text{h}}\Big)}{\text{h}}$
$=\lim_\limits{\text{h}\rightarrow0}\cos\Big(\frac{1}{\text{h}}\Big)$
(LHL at x = c) = (RHL at x = c)
So,
f(x) is differentiable and continuous at x = c.
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